2300 years ago,Moriya Suwakowas defeated byYasaka Kanakoin theGreat Suwa War. As the goddess of mountains inGensokyo, she was planning to make a comeback and regain faith among humans.
To achieve her great plan, she decides to build shrines in human villages. Of course, each village can bulid at most one shrine. If she builds a shrine in thei-th village, she can getFifaith points.
Because of geological differences and the Quantum Entanglement theory, each village has one and only one "entangled" villagePi(it is a kind of one-way relationship). IfSuwakobuilds shrines both in thei-th village and thePi-th village, the faith she get from thei-th village will changes byGipoints (it can result to a negative value of faith points). If she only builds a shrine in thePi-th village, but not in thei-th village, the faith she get will not changes.
Now, please helpSuwakoto find out the maximal faith points she can get.
Input
There are multiple test cases. For each test case:
The first line contains an integerN(2 <=N<= 100000) indicates the number of villages inGensokyo. Then followed byNlines, each line contains three integersFi(0 <=Fi<= 100000)Gi(-100000 <=Gi<= 100000)Pi(1 <=Pi<=NandPiwill not point to thei-th village itself).
Output
For each test case, output the maximal faith points thatSuwakocan get.
Sample Input23 -1 22 -2 143 -2 24 -3 32 -1 15 -2 2Sample Output39
题意：在N个村庄选建圣地，如果在仅在PI建，不会损失满意度，如果在PI的儿子节点
也建会损失G满意度，问你选建时的最大满意度
分析：咋一看跟普通的树DP没啥区别，设dp[i][1]：为在第i个点建的最大满意度，dp[i][0]
不在第i个点建的最大满意度，但是存在环，我们来分析这个环的特点，，每个村庄仅有一个"entangled" village
也就是说连边（有向）的时候每个点仅有一个入边，如果存在环的话，只有可能从环上的点开始向环外的点
边，而且不会存在双环。所以我们就在环上选定一点，分选与不选做两次树形DP，取最大值即可。
对于不在环上的点直接拓扑排序排除掉即可。
#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<string>#include<iostream>#include<queue>#include<cmath>#include<map>#include<stack>#include<set>using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )typedef long long LL;const LL INF = (1LL<<50);typedef pair<int,int>pil;const int maxn=1e5+100;struct node{int to,next;int val;}e[maxn];int n,tot;LL dp[maxn][2];int head[maxn],in[maxn];int vis[maxn],a[maxn];void addedge(int u,int v,int w){e[tot].to=v;e[tot].next=head[u];e[tot].val=w;head[u]=tot++;}void dfs(int u,int fa){dp[u][0]=0;dp[u][1]=a[u];for(int i=head[u];i!=-1;i=e[i].next){int to=e[i].to;if(to!=fa)dfs(to,fa);dp[u][0]+=max(dp[to][0],dp[to][1]);dp[u][1]+=max(dp[to][0],dp[to][1]+e[i].val);}}LL solve(int u){LL ans1=0,ans2=0;dp[u][0]=0;dp[u][1]=-INF;for(int i=head[u];i!=-1;i=e[i].next){int v=e[i].to;dfs(v,u);ans1+=max(dp[v][0],dp[v][1]);}dp[u][0]=-INF;dp[u][1]=a[u];for(int i=head[u];i!=-1;i=e[i].next){int v=e[i].to;dfs(v,u);ans2+=max(dp[v][0],dp[v][1]+e[i].val);}return max(ans1,ans2);}bool ok(int u,int fa){vis[u]=1;for(int i=head[u];i!=-1;i=e[i].next){int to=e[i].to;if(to==fa)return true;if(ok(to,fa))return true;}return false;}void topsort(){queue<int>q;for(int i=1;i<=n;i++)if(!in[i]) q.push(i);while(!q.empty()){int x=q.front();q.pop();vis[x]=1;for(int i=head[x];i!=-1;i=e[i].next){int to=e[i].to;in[to]–;if(!in[to])q.push(to);}}}int main(){int x,y,w;while(~scanf("%d",&n)){CLEAR(head,-1);tot=0;CLEAR(vis,0);CLEAR(in,0);for(int i=1;i<=n;i++){scanf("%d%d%d",&a[i],&w,&y);addedge(y,i,w);in[i]++;}topsort();LL ans=0;for(int i=1;i<=n;i++){if(!vis[i]&&ok(i,i))ans+=solve(i);}printf("%lld\n",ans);}return 0;}
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“过去酒逢知已千杯少，现在酒逢千杯知已少”。
